AMC 10 · 2025 · #25

Grade 11 algebracounting

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

sign-analysispolynomial-rootsdomain-restrictioncombinations-basic work-backwardssystematic-enumeration ↑ Prerequisites: sign-analysispolynomial-roots

📏 Long solution 💡 4 insights

Problem

Two cubic polynomials P(x) and Q(x) are each monic (leading coefficient 1) and every root of each is a number from {1,2,3,4,5}. Their quotient f(x)=

\frac{P(x)}{Q(x)}

is at most 0 on exactly one closed interval [a,b] together with one separate open interval (c,d), where a ＜ b ＜ c ＜ d. Count how many ordered pairs (P,Q) produce this exact solution set.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Work Backwards

We are told the answer set of an inequality and asked to reconstruct the functions, so Tool #11 (Work Backwards) drives everything: read the shape of the solution set to decide where f must vanish and where it must blow up. Tool #4 (Introduce a Variable) names the leftover third roots r and s so the two cubics can be written explicitly. Tool #1 (Draw a Diagram) — a sign chart on the number line — proves the skeleton quotient already produces [a,b]∪(c,d). Tool #7 (Identify Subproblems) splits f into that skeleton times a small leftover factor (x-r)/(x-s) and asks separately what that factor is allowed to do; it forces r=s. Tool #3 (Eliminate Possibilities) then rules out the placements of the shared value r that would break the closed or the open piece, and Tool #2 (Make a Systematic List) walks the five choices of {a,b,c,d} and adds up the survivors.

1STEP 1

Read the endpoints backwards

Outside [a,b] f is positive and the ends are included, so f(a)=f(b)=0 — a,b are roots of P; the open ends c,d must be poles, roots of Q.

f(a)=f(b)=0 → a,b are roots of P; f undefined at c,d → c,d are roots of Q

2STEP 2

Name the leftover roots

Each monic cubic has three roots from {1,2,3,4,5}; P keeps one free root r beyond a,b and Q one free root s beyond c,d.

P(x)=(x-a)(x-b)(x-r), Q(x)=(x-c)(x-d)(x-s), r,s∈{1,2,3,4,5}

3STEP 3

Sign-chart the skeleton

Sign-chart g(x)= $\frac{(x-a)(x-b)}{(x-c)(x-d)}$ : it is negative exactly on (a,b) and (c,d), zero at a,b — so g ≤ 0 is already [a,b]∪(c,d).

g ＜ 0 on (a,b) and (c,d); g=0 at a,b; g ＞ 0 elsewhere

4STEP 4

Force the extra factor to be 1

f=g· $\frac{x-r}{x-s}$ ; the extra factor must never flip a sign, forcing r=s, so it equals 1 everywhere except a hole at x=r.

(x-r)/(x-s) ≥ 0 for all defined x ⇔ r=s, then (x-r)/(x-s)=1 for x ≠ r

5STEP 5

Place the hole so nothing breaks

The hole at x=r must land where the point is already excluded — inside [a,b], (c,d), or at a,b it fails, so r must avoid [a,b]∪(c,d).

r valid ⇔ r∈{1,2,3,4,5}∖([a,b]∪(c,d))

6STEP 6

List the five layouts and add up

Pick 4 of 5 values for a,b,c,d (C(5,4)=5 layouts); valid r counts are 3,2,3,2,3, each fixing one pair, totaling 13.

3+2+3+2+3=13 → (E)

Answer

Test the tightest layout by hand. Take (a,b,c,d)=(1,2,3,4) with r=3: then P(x)=(x-1)(x-2)(x-3) and Q(x)=(x-3)²(x-4), so f(x)=

\frac{(x-1)(x-2)}{(x-3)(x-4)}

away from the hole at x=3. It gives f(1)=f(2)=0, is negative on (1,2) and on (3,4), positive elsewhere, and is undefined at 3 and 4 — exactly [1,2]∪(3,4). A forbidden case confirms the rule too: (1,2,3,4) with r=2 would put a hole at x=2, breaking the closed end of [1,2], so it is correctly excluded. The per-layout counts 3,2,3,2,3 are each at most 3 (only 5 candidate values, at least two always forbidden), so the total 13 is plausible and matches (E).

💡Key takeaway

Closed edges are where f equals zero and open edges are where f blows up; once the skeleton quotient gives the right picture, the only freedom left is where to hide a hole, and counting those holes across the five boundary layouts gives 13 ordered pairs.

Read the endpoints backwards
Name the leftover roots
Sign-chart the skeleton
Force the extra factor to be 1
Place the hole so nothing breaks
List the five layouts and add up